Asked by a colleague: Do we believe that there are an infinite number of number fields that have no unramified extensions? The rational field Q is the most salient example of such a field, and a couple of others are known. But what is conjectured, and what is the philosophy here?

As to the philosophy, note that since such a field has class number 1, this problem is strictly harder than the "there are infinitely many number fields of class number one" problem (which is open, old, and probably very hard).
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Cam McLemanOct 5 '10 at 23:46

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@Mahesh: If we could show that, then for each such $K$ we could consider its maximal unramified extension, say $L$. Then $L$ has no unramified extensions, and so ranging over all $K$, we obtain an infinite number of number fields with no unramified extensions. So your question is equivalent.
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Cam McLemanOct 6 '10 at 1:37

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The question itself is certainly still open. Mostly as an exercise for myself, I'll coalesce my comments above into an answer, and add in some details about where various pieces of the philosophy come from.

The starting point is the following philosophy:

The ring of integers in any number field with sufficiently small root discriminant admits no non-trivial unramified extensions.

This philosophy can occasionally be made precise. For example, Yamamura uses tables of root discriminant bounds from Diaz y Diaz to conclude that for a quadratic imaginary number field $K$ of discriminant $|d|\leq 499$ (or $|d|\leq 2003$ under GRH), the maximal unramified extension of $K$ is a finite extension. This is particularly relevant since each of these maximal unramified extensions clearly has the property that you ask about, that they themselves admit no unramified extensions.

The bad news is that the set of number fields with sufficiently small root discriminant to apply these results (at least, without a tremendous of extra effort analyzing carefully constructed extensions) is finite. In particular, results of Odlyzko imply that that there are only finitely many number fields with root discriminant less than $4\pi e^\gamma\approx 22.3$, where $\gamma$ is the Euler-Mascheroni constant (yeah, that Euler-Mascheroni constant!). Under GRH, this remains true for the larger bound $8\pi e^\gamma$. In fact, as a nice concrete factoid to hold on to, Jones and Roberts have shown that there are exactly 7063 abelian number fields with root discriminant under $8\pi e^\gamma$, and sort these according to their Galois group.

Back to good news: So we now ask ourselves whether or not these numbers $4\pi e^\gamma$ and $8\pi e^\gamma$ can be improved. The answer is a definite yes. If we partition number fields based on their proportion of real and complex embeddings, we can get improvements on fields with increased proportion of complex embeddings, up to an improvement factor of $e$ for totally complex number fields. Further, since Odlyzko's argument stems from work of Stark estimating values of $L$-functions, it seems plausible to believe there are analytic improvements to be made as well. So maybe we can keep pushing these bounds higher and higher, enough so that we find infinitely many number fields with smaller root discriminant.

More bad news: There's an inherent limit to how good we can make these bounds, coming from the study of class field towers. (Okay, so this is actually really good news for those of us who like to study class field towers, but I digress...) Namely, since root discriminants are unchanged when moving up an unramified extension, fields with an infinite class field tower provide a stopping point for any claim of the form "there are finitely many number fields with root discriminant less than such-and-such bound." This is also something that can be partitioned by proportion of real and complex embeddings, and it's been a hot topic recently to see how limited these Odlyzko-type bounds can get. Recently, Hajir and Maire have further refined this line of thought by considering towers of number fields with tame ramification.

So, long story short, from this point of view, the big unknown is whether, once we know optimal bounds on root discriminants, whether or not there will be infinitely many number fields with root discriminants less than that bound. Of course, there's also the possibility that there are other techniques for proving that a number field has no unramified extensions that do not go through root discriminants -- perhaps a form of non-abelian class field theory can come to the rescue, as abelian class field theory can address only the weaker (but still open and fantastically interesting) question of fields with no abelian unramified extensions.

Since you mentioned about the problem of infinitely many fields of class number one, I would like to mention the conjectures that Coates recently made in a talk in Kyoto. Take the extension of $\mathbb{Q}$ with Galois group $\widehat{\mathbb{Z}}$. Then he conjectures that the set class numbers of all fields in this $\widehat{\mathbb{Z}}$ extension is a bounded set. He also conjectures that for any prime $p$ the class number of all fields in the $\mathbb{Z}_p$ extension of $\mathbb{Q}$ is 1. This was conjectured by Weber for $p=2$.
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Mahesh KakdeOct 6 '10 at 11:36

Interesting, thanks. Do you know of a written reference somewhere?
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Cam McLemanOct 6 '10 at 12:28

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No. At this moment it is fair to say that Coates raised them as questions based on some numerical evidence though he called them conjectures. There is a very weak numerical evidence for the second part. Some japanese mathematicians (Okazaki, Fukuda, Komatsu were the names mentioned in the talk of Coates) have shown that for p=2,3 the ideal class groups of fields in the $\mathbb{Z}_p$ extension of $\mathbb{Q}$ has no prime divisor less than a million (I guess).
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Mahesh KakdeOct 6 '10 at 14:20

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I am sorry for a vary late comment on Mahesh' one, but recently Morisawa (student of Komatsu, I think) has shown that given a finite set of primes $S$, for each number field $F$ inside $\mathbb{Q}^{cycl,S}=\prod_{p\in S}\mathbb{Q}^{cycl,p}$, there is a constant $c=c(S,F)$ such that each prime $\ell>c$ whose decomposition field is $F$ does not divide the class number of any other number field contained in $\mathbb{Q}^{cycl,S}$. It appeared in J. Nmber Theory 133 (2013).
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Filippo Alberto EdoardoAug 8 '13 at 2:14